A spring 20cm long is stress to a 25cm by a load of 50n.what will be it new length when stress by 80n assuming that the elastic limit is not exceeded.
We can use the formula for stress and strain to solve this problem:
stress = force / area
strain = change in length / original length
Since the spring is stretched, we can assume that the area is constant. Therefore, we can use the stress to find the strain:
strain = stress / modulus of elasticity
where the modulus of elasticity is a constant that depends on the material of the spring.
Let's assume that the modulus of elasticity for this spring is 20 N/cm². Then, we can calculate the strain caused by the first load of 50 N:
strain = 50 N / (20 N/cm²) = 2.5 cm/cm
This means that the spring increased its length by 2.5 cm for every 1 cm of its original length. Therefore, its new length under a load of 80 N would be:
change in length = strain * original length * force
change in length = (2.5 cm/cm) * (20 cm) * (80 N)
change in length = 400 cm
new length = original length + change in length
new length = 20 cm + 400 cm
new length = 420 cm
Therefore, the new length of the spring under a load of 80 N would be 420 cm, assuming that the elastic limit is not exceeded.
To solve this problem, we can use Hooke's law, which states that the extension of an elastic material is directly proportional to the force applied to it, as long as the elastic limit is not exceeded.
Let's start by finding the spring constant (k) using the given information:
Initial length of the spring, L₁ = 20 cm = 0.20 m
Final length of the spring, L₂ = 25 cm = 0.25 m
Force applied to the spring, F₁ = 50 N
Using Hooke's law: F = kΔL
ΔL = L₂ - L₁
ΔL = 0.25 m - 0.20 m
ΔL = 0.05 m
Substituting these values into Hooke's law:
50 N = k * 0.05 m
Now we can solve for the spring constant, k:
k = 50 N / 0.05 m
k = 1000 N/m
Now that we have the spring constant, we can find the new length of the spring when stressed by 80 N.
Force applied to the spring, F₂ = 80 N
Using Hooke's law: F = kΔL
Rearranging the equation to solve for ΔL:
ΔL = F / k
Substituting the values:
ΔL = 80 N / 1000 N/m
ΔL = 0.08 m
Finally, we can find the new length of the spring:
L = L₁ + ΔL
L = 0.20 m + 0.08 m
L = 0.28 m
Therefore, the new length of the spring when stressed by 80 N is 0.28 m (or 28 cm).
To find the new length of the spring when stressed by 80N, you can use Hooke's Law, which states that the extension of an elastic material is directly proportional to the applied force, as long as the elastic limit is not exceeded.
Hooke's Law equation: F = k * ΔL
where:
F = applied force (in newtons)
k = spring constant (in newtons per meter)
ΔL = change in length (in meters)
First, calculate the spring constant (k) using the given data.
Given:
Original length of the spring (L1) = 20 cm = 0.2 m
Extension of the spring (ΔL) = L2 - L1 = 25 cm - 20 cm = 5 cm = 0.05 m
Applied force (F1) = 50 N
Using Hooke's Law, we can rearrange the equation to solve for the spring constant:
k = F1 / ΔL = (50 N) / (0.05 m) = 1000 N/m
Now that we have the spring constant, we can use it to find the change in length (ΔL2) when stressed by 80N.
Applied force (F2) = 80 N
Rearranging Hooke's Law equation again:
ΔL2 = F2 / k = (80 N) / (1000 N/m) = 0.08 m
To find the new length (L2) when the spring is stressed by 80N:
L2 = L1 + ΔL2 = 0.2 m + 0.08 m = 0.28 m
Therefore, when stressed by 80N, the new length of the spring would be 0.28 meters.